prove that E<sub>p(n!)=[n/p]+[n/p^2]+[n/p^3]+.......till n>=p where n is any natural number, p is any prime number and [n/p] is the greatest integer...
hemajaprove that Ep(n!)=[n/p]+[n/p^2]+[n/p^3]+.......till n>=p
where n is any natural number, p is any prime number and [n/p] is the greatest integer Fungtion of n/p
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